Q: x = 5, y = 2 is a solution of the linear equation

\textbf{Step 1: Check which of the equation passes through (5,2).} \text{The solution should satisfy the appropriate equation.} \text{So, let us investigate each option by substituting }x=5 & y=2 (A) x+2y=7 . \text{L.H.S. }=5+2\times 2=9\neq \text{R.H.S.} \therefore \text{It is not the appropriate equation.} (B) 5x+2y=7 \text{L.H.S. }=5\times 5+2\times 2=29\neq \text{R.H.S.} \therefore \text{It is not the appropriate equation.} (C) x+y=7 \text{L.H.S. }=5+2=7= \text{R.H.S} \therefore \text{It is the appropriate equation.} (D) 5x+y=7 \text{L.H.S. }=5\times 5+2=27\neq \text{R.H.S.} \therefore \text{It is not the appropriate equation.} \textbf{Hence, option C is correct.}

Q: Verify that -(-x)=x for (i) x=\dfrac {11}{15} (ii) x=-\dfrac {13}{17}

i) L.H.S =-(-x) =-\left (-\dfrac {11}{15}\right) = \dfrac {11}{15} R.H.S = x = \dfrac {11}{15} \therefore L.H.S = R.H.S Hence, verified.i) L.H.S =-(-x) =-\left [-\dfrac {(-13)}{17}\right] = -\dfrac {13}{17} R.H.S = x = -\dfrac {13}{17} \therefore L.H.S = R.H.S Hence, verified.

Question 1

Which of the following equations has  x = 2  as a solution?

Accepted Answer

extbf{The value of the variable in an equation which satisfies the equation is called a solution to the} extbf{ equation.} Let us now try and check the value of x=2 for all the four given equationsFor option (A)

x + 2 = 5 2 + 2 \neq = 5

extbf{ Does not satisfy} For option (B)

x - 2 = 0 2 - 2 = 0

extbf{ Satisfies.} For option (C)

2 x + 1 = 0 (2 \times 2) + 1 \neq = 0

extbf{ Does not satisfy} For option (D)

x + 3 = 6 2 + 3 \neq = 6

extbf{ Does not satisfy} extbf{Clearly option B is correct as it gives us the Left hand side equal to the Right hand side.}

Question 2

extbf{State whether the statement is True or False.} 5  is the solution of the equation  3x+2=17 .

Accepted Answer

Consider the given equation,  3 x+2=17 On transposing  2  to RHS.We get, \Rightarrow 3 x=17-2 \Rightarrow 3 x=15 \Rightarrow x=\frac{15}{3} \Rightarrow x=5 	herefore 5 	extbf{ is the solution of the equation } 3 x+2=17 . 	extbf{Hence, the given statement is true.}

Question 3

x = 5, y = 2  is a solution of the linear equation

Accepted Answer

extbf{Step 1: Check which of the equation passes through (5,2).}                     	ext{The solution should satisfy the appropriate equation.}                   	ext{So, let us investigate each option by substituting }x=5  &  y=2                   (A)        x+2y=7 .                  	ext{L.H.S. }=5+2	imes 2=9
eq   	ext{R.H.S.}                    	herefore   	ext{It is not the appropriate equation.}                   (B)        5x+2y=7                   	ext{L.H.S. }=5	imes 5+2	imes 2=29
eq   	ext{R.H.S.}                   	herefore    	ext{It is not the appropriate equation.}                   (C)       x+y=7                    	ext{L.H.S. }=5+2=7=   	ext{R.H.S}                   	herefore   	ext{It  is the appropriate equation.}                   (D)        5x+y=7                    	ext{L.H.S. }=5	imes 5+2=27
eq   	ext{R.H.S.}                   	herefore    	ext{It is not the appropriate equation.} 	extbf{Hence, option C is correct.}

Question 4

x = -1  is a solution of the equation

Accepted Answer

x = -1 2(x-2) + 6 =0 \implies\,  2(-1 -2) + 6 =0 \implies\,  2 	imes (-3) + 6= 0 \implies\,  -6 + 6 =0

Question 5

Which equation among the following don't have  (-1)  as a solution?

Accepted Answer

{	extbf{Step -1: Considering option A.}}                 {	ext{We have given,}}   x +1 = 0                 {	ext{Putting }}x =  - 1{	ext{ we get,}}                  - 1 +1 = 0                  \Rightarrow  0 = 0   {	ext{Which is possible}}{	ext{.}}                 {	ext{Here, L}}{	ext{.H}}{	ext{.S}} = {	ext{R}}{	ext{.H}}{	ext{.S }}                 {	ext{So,  - 1 is a solution of }}x +1 = 0. {	extbf{Step -2: Considering option B.}}                 {	ext{We have given,}}   3x + 4 = 1                 {	ext{Putting }}x =  - 1{	ext{ we get,}}                 3\left( { - 1} ight) + 4 = 1                  \Rightarrow  - 3 + 4 = 1                  \Rightarrow 1 = 1   {	ext{Which is possible}}{	ext{.}}                 {	ext{Here, L}}{	ext{.H}}{	ext{.S }} = {	ext{ R}}{	ext{.H}}{	ext{.S }}                 {	ext{So,  - 1 is a solution of 3}}x + 4 = 1. {	extbf{Step -3: Considering option C.}}                 {	ext{We have given,}}   5x + 7 = 2                 {	ext{Putting }}x =  - 1{	ext{ we get,}}                 5\left( { - 1 } ight) + 7 = 2                  \Rightarrow -5+ 7 = 2                  \Rightarrow  2 = 2   {	ext{Which is possible}}{	ext{.}}                 {	ext{Here, L}}{	ext{.H}}{	ext{.S = R}}{	ext{.H}}{	ext{.S }}                 {	ext{So,  - 1 is a solution of }}5x+7 =2. {	extbf{Step -4: Considering option D.}}                 {	ext{We have given,}}   x -1 = 2                 {	ext{Putting }}x =  - 1{	ext{ we get,}}                 -1-1= 2                  \Rightarrow  - 2 = 2   {	ext{Which is not possible}}{	ext{.}}                 {	ext{Here, L}}{	ext{.H}}{	ext{.S}} 
e {	ext{R}}{	ext{.H}}{	ext{.S }}                 {	ext{So,  - 1 is not a solution of }}x -1 = 2. {	extbf{ Hence, option D. }}\mathbf{ x-1 =2}{	extbf{ is correct answer.}}

Question 6

The solution of the equation  3(x - 4)= 30  is  x =6

Accepted Answer

If  x=6  is the solution then,  3(x-4)=30  is satisfied by  x=6 LHS  =3(6-4)=6 RHS  =30 LHS  
eq  RHS \implies x=6  is not the solution

Question 7

Verify that  x=2  is a solution of the linear equation  2x+7=13-x.

Accepted Answer

Solution: extbf{Step 1: Find the value of L.H.S at } \boldsymbol{x=2} \quad\quad\quad Substitute x=2 in the left side of the given equation, \quad\quad\quad

x + 2 = 5 2 + 2 \neq = 5

extbf{Step 2: Find the value of R.S.H at} \boldsymbol{x=2} \quad\quad\quad Substitute x=2 in the right side of the given equation, \quad\quad\quad

R H S = 13 - 2 = 11

\quad\quad\quad Since, LHS = RHS=11 \ at x=2 . extbf{Hence, 2 is the solution of the given linear equation.}

Question 8

Verify that  -(-x)=x  for (i)  x=\dfrac {11}{15}          (ii)  x=-\dfrac {13}{17}

Accepted Answer

i) L.H.S  =-(-x) =-\left (-\dfrac {11}{15}ight) = \dfrac {11}{15}    R.H.S  = x = \dfrac {11}{15}     	herefore  L.H.S = R.H.S    Hence, verified.i) L.H.S  =-(-x) =-\left [-\dfrac {(-13)}{17}ight] = -\dfrac {13}{17}    R.H.S  = x = -\dfrac {13}{17}     	herefore  L.H.S = R.H.S    Hence, verified.

Question 9

Check whether  p=0  is a solution to  4p-3 = 13  or not.

Accepted Answer

Given:  4p-3= 13 Putting  p=0  in L.H.S., \Rightarrow 4(0)-3=-3 \because L.H.S. 
eq R.H.S.   	herefore p= 0  is not the solution of given equation.

Question 10

Use trial and error method, find the value of  x  if  5x = 25 .

Accepted Answer

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Checking for Solutions of Equation

Important Questions

Which of the following equations has $x = 2$ as a solution?

$State whether the statement is True or False.$
$5$ is the solution of the equation $3 x + 2 = 17$ .

$x = 5, y = 2$ is a solution of the linear equation

$x = - 1$ is a solution of the equation

Which equation among the following don't have $(- 1)$ as a solution?

The solution of the equation $3 (x - 4) = 30$ is $x = 6$

Verify that $x = 2$ is a solution of the linear equation $2 x + 7 = 13 - x .$

Verify that $- (- x) = x$ for
(i) $x = \frac{1 1}{1 5}$ (ii) $x = - \frac{1 3}{1 7}$

Check whether $p = 0$ is a solution to $4 p - 3 = 13$ or not.

Use trial and error method, find the value of $x$ if $5 x = 25$ .

Learn with Videos

Quick Summary With Stories

Checking for Solutions of Equation

Important Questions

Which of the following equations has x=2 as a solution?

State whether the statement is True or False. 5 is the solution of the equation 3x+2=17.

x=5,y=2 is a solution of the linear equation

x=−1 is a solution of the equation

Which equation among the following don't have (−1) as a solution?

The solution of the equation 3(x−4)=30 is x=6

Verify that x=2 is a solution of the linear equation 2x+7=13−x.

Verify that −(−x)=x for (i) x=1511​ (ii) x=−1713​

Check whether p=0 is a solution to 4p−3=13 or not.

Use trial and error method, find the value of x if 5x=25.